These distributions have been computed with a detailed equilibrium model, and they obviously show no sign of a Maxwell/Boltzmann energy distribution throughout the spectrum (their form being mainly determined by radiative recombination and inelastic collisions; see the page Electron Spectrum in the Ionosphere for more information). So in this case (and many others) collisional relaxation towards a Maxwellian velocity distribution simply doesn't take place for some constituents (i.e. the notion of a 'temperature' can strictly speaking not be applied) , and neither can an equipartition of energies between the constituents be assumed to hold here (the latter is especially true if the constituents have very different masses (e.g. electrons and ions), as then the energy exchange for an elastic collision is very small). One could be tempted to conclude from this that if inelastic collisions and dynamic processes are absent (i.e.if only elastic collisions affect the particle energies), the distribution function must necessarily be Maxwellian. This is indeed the conclusion at which Maxwell himself arrived in his 'Dynamical Theory of Gases'. However, in his derivation he makes an assumption which is equivalent to implying a normal distribution (i.e a Gaussian- (Maxwell-) distribution) in the first place (namely by assuming that the initial velocities of the two molecules are perpendicular to each other). Subsequent derivations of the Maxwell velocity distribution using Boltzmann's H-theorem are similarly incorrect in claiming that the Maxwell distribution is not only a sufficient but also a necessary condition for a collisional equilibrium. For instance, Saha in his book 'A Treatise on Heat' (pp.827-833) claims that dH/dt=0 would be synonymous with df/dt=0 as H is merely an integral of f over velocity. It should be clear however that generally ∫F(x)dx=0 does not imply F(x)=0. The conclusion is only correct if one restricts F(x) correspondingly. This is exactly what Saha did in his derivation as he implicitly assumed an isotropic distribution function by assuming the indices of the molecules as inter-changeable. This suggests already that the required condition for the equilibrium distribution being Maxwellian is that it is isotropic, and it is shown in the following that this is indeed the case: assuming a perfectly elastic head-on- (i.e. a 1D-) collision of two particles with masses m1 and m2, the energy- and momentum conservation laws result in the well-known relationship for the velocity v1' of particle 1 after the collision:
(1) v1' = [2m2v2 + (m1-m2)v1]/ (m1+m2) ,
i.e. the kinetic energy of particle 1 after the collision is(2) K1' = m1/2.v1'2 = K1 + 4.m1m2/(m1+m2)2 .[K2-K1 +v1.v2.(m1-m2)/2] ,
where K1 and K2 are the kinetic energies of the particles before the collision. Since energy conservation demands that(3) K1'+K2' = K1+K2 = K = const. ,
it follows that the difference of the kinetic energies after the collision is(4) K2'-K1'= K2-K1 - 8.m1m2/(m1+m2)2 .[K2-K1 +v1.v2.(m1-m2)/2] ,
and if one averages this over the velocity distributions of particles 1 and 2:(5) <K2'>-<K1'>= <K2>-<K1> -8.m1m2/(m1+m2)2 .[<K2>-<K1> +<v1>.<v2>.(m1-m2)/2] ,
where the brackets denote the average.Eq.(5) is the result for a 1D- (head-on) collision. This can easily be generalized to a 3D situation, by noting that the average energy transfer for a 3-D collision is a factor 8/5π/4 smaller than that for a head-on collision (as is evident from a comparison of Eqs.(3) and (11) on my page Coulomb Collision Cross Section and Coulomb Logarithm). Multiplying this value to the factor in front of the square bracket gives thus the corresponding 3D result
(6) <K2'>-<K1'>= <K2>-<K1> -16/5π .m1m2/(m1+m2)2 .[<K2>-<K1> +<v1>.<v2>.(m1-m2)/2] .
It is immediately apparent from Eq.(6) that it depends on the sign of the square bracket on the right hand side of the equation whether the energy difference of the two particles decreases or increases through the collision. Maxwell effectively claimed that the energy difference always decreases, but it is clear from the equation that this is only strictly so if either m1=m2 or <v1>.<v2>=0 (or both). In this case Eq.(6) simplifies to(7) <K2'>-<K1'>= (<K2>-<K1>) .(1 -16/5π .m1m2/(m1+m2)2) ,
and it is obvious that the average energy difference always gets smaller here, i.e. it will asymptotically approach zero in the course of subsequent collisions. In case both velocity distribution functions are isotropic (i.e. <v1>=0 and <v2>=0), these will then approach a Maxwell distribution (with an identical temperature), i.e. collisional relaxation will take place. Even if the initial velocities are not isotropic, it is obvious that they become so after some time if one has a closed system. However, if the system is not closed, and one has significant gain- and loss processes determining the particle distribution function, the latter will not be Maxwellian even in equilibrium (this was effectively already addressed above by mentioning the effect of inelastic collisions, but it could also simply be gains and losses due to gas entering or leaving a given volume). This circumstance is, according to Eq.(7), particularly relevant if the masses m1 and m2 are very different, as then the energy change is very small and it takes a long time for the energy difference to become smaller. In this case, the equilibrium distribution will in general not be given by a Maxwellian. By noting that the condition for a general equilibrium is <K2'>-<K1'>= <K2>-<K1> , this requires according to Eq.(6) merely(8) <K2>-<K1> +<v1>.<v2>.(m1-m2)/2 = 0 ,
which means that, unless m1=m2, the two constituents will have different average kinetic energies in equilibrium if <v1> and <v2> are different from zero. And even if the latter are zero, the kinetic energies may stay different for constituents with much different masses, as then, according to Eq.(7), the energies will change very little in a collision, and collisional equilibrium (through elastic collisions) will not be established. The equilibrium will in this case be determined by other processes (e.g. inelastic collisions and recombination as for the electron spectra shown in the figure above).This shows that the Maxwell distribution is a much stronger assumption than that of an equilibrium in general. The former would in terms of the distribution function f require that
(9) f(K1').f(K2')=f(K1).f(K2)
for which the exponential function would then be the obvious solution because due energy conservation (Eq.(3)) we have (with k denoting the dimensionless, normalized energies K)(10) e-k1'.e-k2' = e-(k1'+k2') = e-(k1+k2) = e-k1.e-k2 .
However, for an equilibrium in general, it is in fact only required that the integrals over the left and right hand side of Eq.(9) are identical, which is a much weaker condition and does not necessarily lead to the solution Eq.(10). Of course, if the system is closed in every sense, then the distribution functions will become isotropic in equilibrium (which essentially corresponds to the assumption of a molecular chaos usually made in gas kinetic theory) and from the above it follows then that they are Maxwellian. Strictly speaking, what one therefore has to do in cases like the one addressed above is to solve the Boltzmann Equation (or appropriate approximations to it) rather than effectively making the assumption of a thermodynamic equilibrium by assuming a Maxwell velocity distribution and applying the notion of a temperature to gaseous constituents. Note: the consideration on this page has very much benefited from an extended email discussion with Dr. P. Radhakrishnamurty